Answer
(a) $24\mu F$
(b) $8.4mA$
Work Step by Step
(a) We can find the required capacitance as follows:
$C=\frac{t}{R}$
We plug in the known values to obtain:
$C=\frac{3.5\times 10^{-3}s}{145\Omega}$
$C=24\mu F$
(b) The required current can be determined as
$I=(\frac{\mathcal{E}}{R})e^{-t/r}$
We plug in the known values to obtain:
$I=(\frac{9.0V}{145\Omega})e^{-\frac{7.0ms}{3.5ms}}$
$I=(0.062A)e^{-2}$
$I=8.4mA$
